Question

根据我所读到的概率，开关门应该有66％的机会选择正确的门。下面这段代码就是我提出来的，它大概有50％的胜利，而不是我期待的66％。任何有关我在哪里出错的帮助将非常感激。

for (int count = 0; count < 10000; count++)
{
    // Chooses which door contains DAT GRAND PRIZE YO.
    wDoor = rand() % 3 + 1;

    // AI Contestants Door choice
    aiDoor = rand() % 3 + 1;

    // Using oldChoice to ensure same door isn't picked.
    oldChoice = aiDoor;
    // Used in determining what door to open.
    openedDoor = aiDoor;

    // "Open" a door that is not the winning door and not the door chosen by player.
    do
    {
                openedDoor = rand() % 3 + 1;

    }while (openedDoor != wDoor && openedDoor != aiDoor);

    // Select new door between the remaining two.
    do
    {
              aiDoor = rand() % 3 + 1;

    }while (aiDoor != oldChoice && aiDoor != openedDoor);

    // Increment win counter if new door is correct.
    if (aiDoor == wDoor)
    {
               chooseAgain++;
    }

}

Answer 1

您的while条件是错误的：

while (openedDoor != wDoor && openedDoor != aiDoor)

应该是

while (openedDoor == wDoor || openedDoor == aiDoor)

等

Answer 2

你的条件已经逆转了。 do ... while（...）循环将按照您的注释描述执行，如果它们重复...直到（...），它具有与终止测试相反的极性。

取消实现所需算法的条件。

请注意，在这两种情况下，您最多有两扇门可供选择。使用这些知识，您可以确定隔壁最多只使用一次rand（）而没有循环。

Answer 3

// "Open" a door that is not the winning door and not the door chosen by player.
    do
    {
                openedDoor = rand() % 3 + 1;

    }while (openedDoor != wDoor && openedDoor != aiDoor);

当您打开获胜门（！）或玩家选择的门时，此条件为假（即循环结束）。这与你想要的相反。

    // Select new door between the remaining two.
    do
    {
              aiDoor = rand() % 3 + 1;

    }while (aiDoor != oldChoice && aiDoor != openedDoor);

当玩家选择与之前相同的门或打开的门时，这种情况是错误的（即循环结束）。这也与你想要的相反。

反转条件会得到预期的结果（~0.66）。

蒙蒂霍尔模拟中的意外结果

3 个答案: